Question 15.11: Determine the deflection of the mid-span point of the linear...
Determine the deflection of the mid-span point of the linearly elastic, simply supported beam ABC shown in Fig. 15.23(a).
We shall suppose that the deflected shape of the beam is unknown. Initially, therefore, we shall assume a deflected shape that satisfies the boundary conditions for the beam. Generally, trigonometric or polynomial functions have been found to be the most convenient where the simpler the function the less accurate the solution. Let us suppose that the displaced shape of the beam is given by
v=v_{\mathrm{B}} \sin \frac{\pi x}{L} (i)
in which v_{\mathrm{B}} is the deflection at the mid-span point. From Eq. (i) we see that when x=0 and x=L, v=0 and that when x=L / 2, v=v_{\mathrm{B}}. Furthermore, \mathrm{d} v / \mathrm{d} x= (\pi / L) v_{\mathrm{B}} \cos (\pi x / L) which is zero when x=L / 2. Thus the displacement function satisfies the boundary conditions of the beam.
The strain energy due to bending of the beam is given by Eq. (9.21),
U=\int_{L}{\frac{M^{2}}{2E I_{z}}}\,\mathrm{d}x (9.21)
i.e.
U=\int_{0}^{L} \frac{M^{2}}{2 E I} \mathrm{~d} x (ii)
Also, from Eq. (13.3)
{\frac{\mathrm{d}^{2}v}{\mathrm{d}x^{2}}}={\frac{M}{E I}} (13.3)
M=E I \frac{\mathrm{d}^{2} v}{\mathrm{~d} x^{2}} (iii)
Substituting in Eq. (iii) for v from Eq. (i), and for M in Eq. (ii) from Eq. (iii), we have
U=\frac{E I}{2} \int_{0}^{L} \frac{v_{\mathrm{B}}^{2} \pi^{4}}{L^{4}} \sin ^{2} \frac{\pi x}{L} \mathrm{~d} x
which gives
U=\frac{\pi^{4} E I v_{\mathrm{B}}^{2}}{4 L^{3}}
The TPE of the beam is then given by
\mathrm{TPE}=U+V=\frac{\pi^{4} E I v_{\mathrm{B}}^{2}}{4 L^{3}}-W v_{\mathrm{B}}
Hence, from the principle of the stationary value of the TPE
\frac{\partial(U+V)}{\partial v_{\mathrm{B}}}=\frac{\pi^{4} E I v_{\mathrm{B}}}{2 L^{3}}-W=0
whence
v_{\mathrm{B}}=\frac{2 W L^{3}}{\pi^{4} E I}=0.02053 \frac{W L^{3}}{E I} (iv)
The exact expression for the deflection at the mid-span point was found in Ex. 13.5 and is
v_{\mathrm{B}}=\frac{W L^{3}}{48 E I}=0.02083 \frac{W L^{3}}{E I} (v)
Comparing the exact and approximate results we see that the difference is less than 2 \%. Furthermore, the approximate deflection is less than the exact deflection because, by assuming a deflected shape, we have, in effect, forced the beam into that shape by imposing restraints; the beam is therefore stiffer.